Prove that product of three consecutive integers is divisible by 6.

The product of three consecutive integers which are divisible by 6 is:

⟹6 /n(n+1)(n+2)

We have to prove it by mathematical induction

CASE-I

For n=1

= n(n+1)(n+2) as we have

Putting n=1 into above, we have

=1(1+1)(1+2)

=1(2)(3)

CASE-I is true for n=1

CASE-II

Suppose it is true for n=k, ∀ k ∈ Z

i.e 6/k(k+1)(k+2) says Eq. (1)

We will prove it for n=k+1, we have

n(n+1)(n+2) = (k+1)(k+1+1)(k+1+2)

n(n+1)(n+2) = (k+1)(k+2)(k+3)

n(n+1)(n+2) = k(k+1)(k+2) + 3(k+1)(K+2) says Eq. (2)

Now one of them is k+1 , k+2 will be even. So (k+1)(k+2) will be even.

⇒2 /(k+1)(k+2)

Multiplying both sides with ‘3’, we have

⇒3.2 / 3(k+1)(k+2)

⇒ 6 / 3(k+1)(k+2) says Eq. (3)

From Eq. (1) & Eq. (3), we have

⇒ 6 / k(k+1)(k+2) + 3(k+1)(k+2)

From Eq. (2), we have

⇒ 6 / n(n+1)(n+2)

⇒ CASE-II is also true for n=k+1

Hence proved that product of three consecutive integers is divisible by 6.

Prove that product of three consecutive integers is divisible by 6.

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